Question

Let $A$ and $B$ be any two points on the lines represented by $4x^2-9y^2=0.$ If the area of triangle $OAB$ is $5$ $\left(O\text{is origin}\right)$ then which of the following is the possible equation of the locus of midpoint of $AB?$

• A. $4x^2+9y^2+30=0$
• B. $4x^2-9y^2-30=0$
• C. $9x^2-4y^2-30=0$
• D. $9x^2+4y^2-30=0$

Answer 1

The correct option is B $4x^2-9y^2-30=0$


$\text{Area of}△OAB=|\frac{1}{2}OA.OB\sin \theta |$
$=5$

$OA=\frac{\sqrt{13}}{3}h;OB=\frac{\sqrt{13}}{3}k$

$5=|\frac{1}{2}.\frac{\sqrt{13}}{3}h.\frac{\sqrt{13}}{3}k.\sin \theta |$

$\tan \theta =\frac{\frac{2}{3}-(-\frac{2}{3})}{1+\frac{2}{3}-(-\frac{2}{3})}=\frac{12}{5}\Rightarrow \sin \theta =\frac{12}{13}$

$\Rightarrow |hk|=\frac{15}{2}\Rightarrow hk=\pm \frac{15}{2}$

Midpoint of $AB=(\frac{h+k}{2},\frac{2\frac{h}{3}-2\frac{k}{3}}{2})$

$=(\frac{h+k}{2},\frac{h-k}{3})$
$\Rightarrow h+k=2x;h-k=3y$
$\Rightarrow 2h=2x+3y$
$\Rightarrow 2k=2x-3y$
$\Rightarrow 4hk=4x^2-9y^2$
$\Rightarrow 4x^2-9y^2=\pm 30$

Alternate Solution:
$\text{Area of}△OAB=\frac{1}{2}\left|\begin{array}{ccc}1& 0& 0\\ 1& k& -\frac{2k}{3}\\ 1& h& \frac{2h}{3}\end{array}\right|$
$\Rightarrow 5=\frac{1}{2}|\frac{2kh}{3}+\frac{2kh}{3}|$
$\Rightarrow |hk|=\frac{15}{2}\Rightarrow hk=\pm \frac{15}{2}$

Midpoint of $AB=(\frac{h+k}{2},\frac{2\frac{h}{3}-2\frac{k}{3}}{2})$

$=(\frac{h+k}{2},\frac{h-k}{3})$
$\Rightarrow h+k=2x;h-k=3y$
$\Rightarrow 2h=2x+3y$
$\Rightarrow 2k=2x-3y$
$\Rightarrow 4hk=4x^2-9y^2$
$\Rightarrow 4x^2-9y^2=\pm 30$